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I'm pretty new to the idea of tensors, and I'm having a bit of confusion with how to think about the flat space metric tensor in special relativity.

I understand that a good way to think about tensors is as multilinear functions, which takes a number of vectors from a vector space $V$ over a field $F$ as arguments, and returns a scalar from $F$.

However, I don't really know how to think about the flat space / Minkowski metric tensor in this way. In my class, I understand it's come up in the equation

$ds^2 = \eta_{\alpha \beta} dx^\alpha dx^\beta$

where $\eta_{\alpha \beta}$ is the metric tensor, and $dx^\alpha$ is the $\alpha$'th component of the differential 4-vector. I just don't really see how we can think of $\eta$ as a multilinear function here. Can anyone shed any light on this?

Thanks!

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  • $\begingroup$ The equation you wrote is equivalent to $ds^2 = \eta(dx, dx)$. So you see that the metric is a function that takes in two arguments and returns a number, and furthermore from the equation you wrote, it is linear in each argument. $\endgroup$ – knzhou Sep 28 at 20:18
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This longer answer is a supplement to tparker's good concise answer.


The notation $dx^a$ is used for a couple of different-but-related things:

  • First perspective: Intuitively, we often think of it as an infinitesimal displacement. In this view, $ds^2=g_{ab}dx^a dx^b$ is giving us the infinitesimal change $ds$ in proper time or proper distance (depending on the sign convention) along an infinitesimal segment of a worldline. In this perspective, the equation $ds^2=g_{ab}dx^a dx^b$ is really an abbreviation for $$ \left(\frac{ds}{d\lambda}\right)^2 =g_{ab}\frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda} $$ where $\lambda$ is a parameter that runs along a given worldline so that the $\lambda$th point has coordinates $x^a(\lambda)$.

  • Second perspective: More mathematically, $dx^a$ is a one-form, which takes a vector field as input and returns its $a$th component as output. In terms of coordinates, a vector field $V$ is a linear combination of partial derivatives, $V=V^a\partial_a$, and the one-form $dx^a$ has the property $dx^a(V) = V^a$. Equivalently, $dx^a(\partial_b)=\delta^a_b$.

The second perspective most directly explains why people say that the metric is a bilinear function. The metric $g=g_{ab}dx^a dx^b$ takes two vectors fields as input and returns a scalar field as output, like this (see tparker's answer): $$ g(A,B) = g_{ab}dx^a(A) dx^b(B) = g_{ab} A^a B^b. $$ To relate this to the first perspective, think of the given worldline as one member of a congruence of non-overlapping worldlines that covers the whole spacetime (within a given region), all parameterized by $\lambda$ in such a way that $\lambda$ can be viewed as a new coordinate in some new coordinate system. The vector field $\partial/\partial\lambda$ is everywhere tangent to those worldlines. In terms of the original coordinate system, this vector field is $$ \frac{\partial}{\partial\lambda} = \frac{dx^a}{d\lambda} \frac{\partial}{\partial x^a}, $$ so its components are $$ dx^a\left(\frac{\partial}{\partial\lambda}\right) = \frac{dx^a}{d\lambda}. $$ Use this to get $$ g\left(\frac{\partial}{\partial\lambda}, \,\frac{\partial}{\partial\lambda}\right) =g_{ab}\frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda}, $$ which shows the relationship between the two perspectives. Note the two different usages of the notation $dx^a$, sometimes as a one-form and sometimes as the numerator of a derivative (that is, as an infinitesimal displacement).

This all applies in flat spacetime just as it does in curved spacetime, because this all applies whether or not the components $g_{ab}$ are themselves functions of the coordinates.

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The metric $\eta$ is a bilinear function that inputs two four-vectors (call them $A$ and $B$) and outputs the scalar $A \cdot B = A^\mu B_\mu = A^\mu \eta_{\mu \nu} B^\nu$. It is clearly bilinear (i.e. separately linear in both the first argument $A$ and the second argument $B$) by the distributive property.

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